Consider all the different possible combinations of 4 members of the committee (b,b,b,b), (b,b,b,g),...(g,g,g,g). We need to use the binomial distribution given below
[tex]P(k)=(nbinomialk)p^k(1-p)^{n-k}[/tex]In our case
[tex]k=2,n=4,p=\frac{10}{10+12}=\frac{10}{22}=\frac{5}{11}[/tex]Then,
[tex]\begin{gathered} P(2)=(\frac{4!}{2!(4-2)!})(\frac{5}{11})^2(\frac{6}{11})^2 \\ \Rightarrow P(2)=6\cdot\frac{900}{14641} \\ \Rightarrow P(2)=0. \end{gathered}[/tex]